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ISRAEL JOURNAL OF MATHEMATICS 194 (2013), 69–76
DOI: 10.1007/s11856-012-0078-0
BOUNDS ON THE NUMBER OF AUTOMORPHISMS OFCURVES OVER ALGEBRAICALLY CLOSED FIELDS
BY
Barry Green
African Institute for Mathematical Sciences
Muizenberg, 9745 Cape Town, South Africa
and
Department of Mathematical Sciences, University of Stellenbosch
7602 Stellenbosch, South Africa
e-mail: [email protected]
ABSTRACT
In this note we compare the maximum number of automorphisms of smooth
projective curves of genus g ≥ 2 over algebraically closed fields of charac-
teristic 0 or a prime p > 0, respectively. For given g ≥ 2, we show that for
almost all primes p > 0, this number is the same.
1. Introduction
The aim of this note is give partial answers to questions posed by Oort in [O]
on the group of automorphisms of smooth projective curves over algebraically
closed fields. We investigate the bounds on the possible orders such groups
can have and compare what is known in characteristic zero with questions and
conjectures in positive characteristic, some of which are answered here.
Let K be an algebraically closed field and let C/K be a smooth projective
curve with genus gC , which is assumed greater than or equal to 2 throughout
this note. We are interested in #(AutK(C)) and recall that if char(K) = 0,
then by results of Hurwitz, #(AutK(C)) ≤ 84(gC − 1) and that this is also true
if char(K) = p > 2gC + 1. See for example [Sa], p. 582, or [R2] for a stronger
result with one exception.
Received July 12, 2011
69
70 BARRY GREEN Isr. J. Math.
We are interested in the lower and upper bounds of
μ(g, char(K)) := maxgC=g
#((AutK(C)),
and note that if char(K) = 0 then results of Macbeath, [M], Maclachlan, [Mc],
and Accola, [A1], prove that
8(g + 1) ≤ μ(g, 0) ≤ 84(g − 1)
with each of the bounds being attained for infinitely many values of g. The
result on the upper bound was demonstrated in [M] and for the lower bounds
in [Mc] and [A1].
1.1. Question (Oort) [O]): If p is any prime and K is an algebraically closed
field with char(K) = p, does there exist a non-zero polynomial Mp(T ) ∈ Q[T ]
such that:
(i) For all g ≥ 2, Mp(g) ≤ μ(g, p).
(ii) For infinitely many g, Mp(g) = μ(g, p).
Furthermore, if such a polynomial exists, is it the same for different values of
p and is it equal to 8(g + 1)?
1.2. We prove:
(i) Given g ≥ 2, for almost all primes p > 0, μ(g, p) ≥ 8(g + 1).
(ii) For each positive integer m there exist integers 2 ≤ g1 < g2 < · · · < gm
and 2 ≤ g′1 < g′2 < · · · < g′m, such that for 1 ≤ i ≤ m :
μ(gi, p) = μ(gi, 0) = 8(gi + 1) for almost all p > 84(gi − 1) and
μ(g′i, p) = μ(g′i, 0) = 84(g′i − 1) for all p > 84(g′i − 1).
For a fixed prime p these results don’t answer Oort’s question, but we
will prove a weaker result, namely:
(iii) Let g ≥ 2 with μ(g, 0) = 8(g + 1). Then for almost all primes p > 0,
μ(g, p) = 8(g + 1).
Acknowledgement. I would like to thank Moshe Jarden for helpful sugges-
tions for improving this note.
Vol. 194, 2013 BOUNDS ON THE AUTOMORPHISM GROUP OF CURVES 71
2. Good reduction of curves and their automorphism groups
Let K be an algebraically closed field of characteristic zero and M be the set of
non-archimedean valuations on K of mixed characteristic. For each v ∈ M let
Kv denote the residue field, which has positive characteristic, p say. Let C/K
be a smooth projective irreducible curve of genus gC ≥ 2. Then for almost all
primes p, there exist v ∈ M of residue characteristic p, such that C has good
reduction at v; see for example [D], [R] or [GMP I]. We shall denote the reduced
curve by Cv/Kv, which is smooth, projective and has genus gCv = gC .
In this situation there is a naturally well defined homomorphism
Φ:AutK(C) −→ AutKv(Cv).
For convenience, we recall how this homomorphism is constructed: Let F =
κ(C) denote the function field of C, and recall that the reduction of the curve
C at v corresponds to a valuation prolongation of v on K to F (called a con-
stant reduction), which we also denote by v. We show that the action of the
automorphisms of F |K on this valuation v is invariant. If τ ∈ Aut(F |K), then
w = v ◦ τ−1 is a valuation of F |K whose valuation ring and maximal ideals are
τ(Ov) and τ(Mv), respectively. Let Fv, respectively Fw, denote the residue
fields of F with respect to v, respectively w. It follows directly that the map
τ : Fv → Fw := τ(Ov)/τ(Mv)
defined by fv �→ τ(f) + τ(Mv) is a well-defined isomorphism that fixes each
element of Kv. Hence the function fields Fv|Kv and Fw|Kv have the same
genus. We deduce that w = v ◦ τ−1 is also a good reduction of F |K and by
the uniqueness of the good reduction over a given valuation of K (see [GMP I],
Corollary 3.2), it follows that w = v, so Fw = Fv. Hence we have a well defined
homomorphism
Ψ:Aut(F |K) −→ Aut(Fv|Kv).
Observing that κ(Cv) = Fv this map induces the homomorphism Φ above.
Theorem 2.1 ([K]): Let Φ : AutK(C) −→ AutKv(Cv) be as defined above.
Then:
(a) Φ is injective.
72 BARRY GREEN Isr. J. Math.
(b) For each G ⊂ AutK(C), the restriction ΦG : G −→ AutKv(Cv/Dv) is
surjective, where D := C/G.1
The proofs of these results can be found in [K] in the context of function
fields and their constant reductions. As we shall be using (a) in what follows
we include a short proof here for our context and notation.
Proof of (a). Suppose σ ∈ KerΦ with σ = 1C , and by abuse of notation we
denote its action on the function field F = κ(C) by σ too. We first observe
that σ ∈ ker(Φ) if and only if σ(x) − x ∈ Mv for each x ∈ Ov ⊂ F. Therefore,
ker(Φ) is the inertia group, say T, of the the extension F |E, where E = FG
and G = Aut(F |K) ≈ AutK(C). Now as F |K has good reduction at v it
satisfies, v(F×) = v(K×), so v(F×) = v(E×), [GMP I], Corollary 3.2. By [ZS],
Chapter VI, §12, there is a homomorphism of the inertia group T of F |E into
Hom(v(F×)/v(E×), (Fv)×) whose kernel is the ramification group V. It follows
that T = V . It is well known (see for example [ZS], Theorem 24, p. 77) that this
group is a p-Sylow subgroup and Fv|Ev is a purely inseparable extension. We
have shown that KerΦ is a p-group. (See also [E], Chapter III, or [Ef], Chapter
16, for these facts.)
We next examine the covering C → D := C/〈σ〉 and its reduction with
respect to v, Cv → Dv. Then as 〈σ〉 is a p-group both coverings determine p
extensions of the function fields, with the latter being purely inseparable as Kv
has characteristic p. Now since the extension of function fields determined by Cv
and Dv is purely inseparable, it follows that these fields are isomorphic (see for
example [St], Proposition 3.10.2) and so they have the same genus. Therefore
Cv and Dv have the same geometric genus, say g, which is equal to gC as C
has good reduction at v.
Noting that the genus of D is greater than or equal to the geometric genus
of its reduction Dv, we obtain the following inequality of genera:2
gC ≥ gD ≥ g = gCv = gC .
Hence it follows that we have equalities and so D has good reduction at v and
gC = gD. Finally, by the Riemann–Hurwitz formula and since gC ≥ 2, this
1 Although not needed in this note, we remark that as C/K has good reduction at v, so
does D/K. See for example [Y] and [G].2 See for example [GMP I], Corollary 2.5, for a proof of this fact in the context of constant
reductions.
Vol. 194, 2013 BOUNDS ON THE AUTOMORPHISM GROUP OF CURVES 73
can only happen if C = D and consequently σ = 1C , a contradiction. This
completes the proof of (a).
Question: Under which conditions is Φ surjective?
Theorem 2.2 below gives a partial answer to this question:
Theorem 2.2: Theorem 2.2. Let C/K and Cv/Kv be as above and suppose
the order of G = AutKv(Cv) is relatively prime to p. Then Φ is an isomorphism.
Proof. By Grothendieck (SGA 1, Expose XIII, §2) for C/K as above we can
lift G to G ⊂ AutK(C) such that G ∼= G. Since Φ is injective the result follows.
See also [CGH], sections 1 and 2.
Corollary 2.3: Let C/K be a smooth projective irreducible curve which has
good reduction at v ∈ M. If char(Kv) = p > 84(gC − 1)), then Φ is an isomor-
phism.
Proof. Since C/K has good reduction at v and p > 84(gC − 1), we have
p > 84(gCv − 1) > 2gCv − 1. Hence, by the second paragraph of the intro-
duction it follows that
#(AutKv(Cv)) ≤ 84(gCv − 1).
Therefore (#(AutKv(Cv)), p) = 1 and, by Theorem 2.2, Φ is an
isomorphism.
Corollary 2.4: Let g ≥ 2 be given. Then for all primes p > 84(g − 1),
μ(g, 0) ≥ μ(g, p). If in addition μ(g, p) = 84(g − 1), then μ(g, 0) = μ(g, p).
Proof. Let K be an algebraically closed field of characteristic p and C/K be a
smooth projective irreducible curve with genus g and #(AutK(C)) = μ(g, p).
Next suppose (C/K, v) is a good lifting of C/K to characteristic 0, with K
algebraically closed and Kv = K. Then as p > 84(g − 1), Theorem 2.2 implies
that Φ is an isomorphism and μ(g, 0) ≥ #(AutK(C)) = #(AutK(C)) = μ(g, p).
The final assertion follows directly from the upper bound on the number of
automorphisms in characteristic 0.
We next recall a general fact on the existence of good reductions of smooth
projective irreducible curves C/K. For a proof see [D], [R1], or the observation
[GMP I], 1.3.
74 BARRY GREEN Isr. J. Math.
2.5. Deuring/Roquette. Let C/K be a smooth projective irreducible curve
and suppose F = κ(C). Then for each non-constant f ∈ F, there exists a finite
set A ⊂ K such that for each v ∈ M, if v(A) ≥ 0, then the Gauss valuation vf
on K(f) has a unique extension to F, which is a good reduction of C/K. Hence
for almost all primes p, there exists v ∈ M of residue characteristic p, such that
C has good reduction at v.
Proposition 2.6: Let g ≥ 2 be given. Then for almost all primes p, μ(g, 0) ≤μ(g, p). If in addition p > 84(g− 1) then for almost all such p, μ(g, 0) = μ(g, p).
Proof. Given g ≥ 2, let C/K be chosen so that gC = g and #(AutK(C)) =
μ(g, 0). Then by 2.5 above, for almost all primes p there exists v defined on K
so that char(Kv) = p and C/K has good reduction at v. It follows by Theorem
2.1 that μ(g, 0) = #(AutK(C)) ≤ #(AutKv(Cv)) ≤ μ(g, p). The final assertion
follows by combining this result with 2.4 above.
We return to Oort’s question:
3. Bounds on the number of automorphisms
We now prove the results stated in 1.2.
Proposition 3.1: Let g ≥ 2 be given. Then for almost all primes p > 0,
μ(g, p) ≥ 8(g + 1).
Proof. By Proposition 2.6 for almost all p > 0, μ(g, p) ≥ μ(g, 0). Therefore
since μ(g, 0) ≥ 8(g + 1) for all g ≥ 2, the result follows.
Applying Theorem 2.1 we prove 1.2 (iii) from the introduction. For conve-
nience we first define S := {g ∈ N | μ(g, 0) = 8(g + 1) }, which is an infinite set
by [Mc], Theorem 5.
Proposition 3.2: Let g ∈ S. Then for almost all primes p > 0, μ(g, p) =
μ(g, 0) = 8(g + 1).
Proof. This follows directly from Proposition 2.6.
Theorem 3.3: For each positive integer m there exist integers
2 ≤ g1 < g2 < · · · ≤ gm and 2 ≤ g′1 < g′2 < · · · < g′m,
such that for 1 ≤ i ≤ m,
Vol. 194, 2013 BOUNDS ON THE AUTOMORPHISM GROUP OF CURVES 75
(i) μ(gi, p) = μ(gi, 0) = 8(gi + 1) for almost all primes p > 84(gi − 1),
(ii) μ(g′i, p) = μ(g′i, 0) = 84(g′i − 1) for all primes p > 84(g′i − 1).
Proof. (i) For m ∈ N, choose an increasing sequence g1, g2, . . . , gm ∈ S. Then
for each i, 1 ≤ i ≤ m, by 2.6, μ(gi, p) = μ(gi, 0) = 8(gi+1) for almost all primes
p > 84(gi − 1).
(ii) For m ∈ N, choose an increasing sequence
g′1, g′2, . . . , g
′m ∈ T := {g′ ∈ N | μ(g′, 0) = 84(g′ − 1) },
which is infinite and by [M], Corollary to Theorem 6. Then ii) follows immedi-
ately from 2.4 for all primes p > 84(g′i − 1).
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